MathGroup Archive 2004

[Date Index] [Thread Index] [Author Index]

Search the Archive

pair functions applied to Zeta function in two ways


When I thought of this , I felt that it might work for the zeta zeros,
but experience with it in Mathematica shows these sums functions are
too far from analytical in s=1/2 strip.

(* pair functions applied to Zeta function in two ways*)
(*the first with {1/(n+1),n/(1+n)} as the pair gives nearly a straight 
line as parametricPlot*)
(* second with {1/(x+1),x/(1+x)} as the pair gives seems like conic 
curve:parabolic or hyperbolic*)
(* The second has a definite minimum due to the gzetax near 3.5*)
(* it is doubtful these sums will be any good in the zeta zero region of 
1/2+I*b[n]*)
(* where fzeta[1/2+I*b[n]]=-gzeta[1/2+I*b[n]]: 
fzetax[1/2+I*b[n]]=-gzetax[1/2+I*b[n]] *)

fzeta[x_]=Sum[If[Mod[n,2]==1,(1/(n+1)),(n/(n+1))]/n^x,{n,1,Infinity}]
gzeta[x_]=Sum[If[Mod[n,2]==1,(n/(n+1)),(1/(n+1))]/n^x,{n,1,Infinity}]
ParametricPlot[{N[fzeta[x]],N[gzeta[x]]},{x,2,5}]
Plot[N[fzeta[x]],{x,2,5},PlotRange->All]
Plot[N[gzeta[x]],{x,2,5},PlotRange->All]
fzetax[x_]=Sum[If[Mod[n,2]==1,(1/(x+1)),(x/(x+1))]/n^x,{n,1,Infinity}]
gzetax[x_]=Sum[If[Mod[n,2]==1,(x/(x+1)),(1/(x+1))]/n^x,{n,1,Infinity}]
ParametricPlot[{N[fzetax[x]],N[gzetax[x]]},{x,2,5},PlotRange->All]
Plot[N[fzetax[x]],{x,2,5},PlotRange->All]
Respectfully, Roger L. Bagula

tftn at earthlink.net, 11759Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 :
alternative email: rlbtftn at netscape.net
URL :  http://home.earthlink.net/~tftn


  • Prev by Date: Re: canonical mathematical expression represenation?
  • Next by Date: Re: Proving inequalities with Mathematica
  • Previous by thread: help on sublists xor
  • Next by thread: New User - Programming